Let alpha and beta be the roots of x^2 - 6x - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 - 6x - 2 = 0$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$\alpha^2 - 6\a

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 - 6x - 2 = 0$.Since $\alpha$ and $\beta$ are roots,they satisfy the equation:$\alpha^2 - 6\alpha - 2 = 0 \implies \alpha^2 - 2 = 6\alpha$$\beta^2 - 6\beta - 2 = 0 \implies \beta^2 - 2 = 6\beta$We need to evaluate $\frac{a_{10} - 2a_8}{2a_9}$.Substituting $a_n = \alpha^n - \beta^n$:$\frac{(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)}{2(\alpha^9 - \beta^9)}$$= \frac{\alpha^8(\alpha^2 - 2) - \beta^8(\beta^2 - 2)}{2(\alpha^9 - \beta^9)}$Using the relations $\alpha^2 - 2 = 6\alpha$ and $\beta^2 - 2 = 6\beta$:$= \frac{\alpha^8(6\alpha) - \beta^8(6\beta)}{2(\alpha^9 - \beta^9)}$$= \frac{6\alpha^9 - 6\beta^9}{2(\alpha^9 - \beta^9)}$$= \frac{6(\alpha^9 - \beta^9)}{2(\alpha^9 - \beta^9)} = \frac{6}{2} = 3$.

Let $\alpha$ and $\beta$ be the roots of $x^2 - 6x - 2 = 0$,where $\alpha > \beta$. If $a_n = \alpha^n - \beta^n$ for all $n \geq 1$,then what is the value of $\frac{a_{10} - 2a_8}{2a_9}$?

Similar Questions

In a triangle $PQR$, $\angle R = \pi / 2$. If $\tan(P/2)$ and $\tan(Q/2)$ are the roots of the quadratic equation $ax^2 + bx + c = 0$, where $a \neq 0$, then which of the following is true?

If $\alpha, \beta, \gamma$ are the roots of the equation $2x^3+x^2-13x+6=0$,then $\alpha^3+\beta^3+\gamma^3=$

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^{3}+4x+2=0$,then $\alpha^{3}+\beta^{3}+\gamma^{3}$ is equal to

If $\alpha, \beta$ are the roots of the quadratic equation $x^{2}+a x+b=0, (b \neq 0),$ then the quadratic equation whose roots are $\alpha-\frac{1}{\beta}$ and $\beta-\frac{1}{\alpha}$ is:

If the harmonic mean of the roots of the equation $\sqrt{2} x^2 - bx + (8 - 2\sqrt{5}) = 0$ is $4$,then the value of $b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module